Question

Помогите решить пожалуйста, срочно нужно вычисления пределов.​ lim x-0

Answer 1

$1) ~ \displaystyle \lim_{x \to \infty} \frac{1 - \cos x}{\text{tg} \, x^{2}} = \left | {{x = \dfrac{1}{t} ; ~ t = \dfrac{1}{x} } \atop {x \to \infty; ~ t \to 0}} \right | = \lim_{t \to 0} \frac{1 - \cos \dfrac{1}{t}}{\text{tg} \, \dfrac{1}{t^{2}}} =\lim_{t \to 0} \frac{2\sin^{2} \dfrac{1}{2t} }{\text{tg} \, \dfrac{1}{t^{2}}} =$

$\displaystyle = \left|\begin{array}{ccc}\sin \dfrac{1}{2t} \sim \dfrac{1}{2t} \\ \\\text{tg} \, \dfrac{1}{t^{2}} \sim \dfrac{1}{t^{2}} \\ \\t \to 0\end{array}\right| = \lim_{t \to 0} \dfrac{2 \cdot \left(\dfrac{1}{2t} \right)^{2}}{\dfrac{1}{t^{2}} } = \lim_{t \to 0} \dfrac{2 \cdot \dfrac{1}{4t^{2}} }{\dfrac{1}{t^{2}} } = \dfrac{1}{2}$

$2) ~ \displaystyle \lim_{x \to 0} \frac{\text{tg}^{3} \, 2x}{e^{2x^{3}} - 1} =\left[\frac{0}{0} \right] = \left|\begin{array}{ccc}\text{tg} \, 2x \sim 2x \\e^{2x^{3}} - 1 \sim 2x^{3}\\x \to 0\end{array}\right| =\lim_{x \to 0} \frac{(2x)^{3}}{2x^{3}} = \lim_{x \to 0} \frac{8x^{3}}{2x^{3}} = 4$